Q#19.47
Figure P19.47 shows a pV-diagram for 0.0040 mole of ideal gas.
The temperature of the gas does not change during segment bc.
(a) What volume does this gas occupy at point c?
(b) Find the temperature of the gas at points a, b, and c.
(c) How much heat went into or out of the gas during segments ab, ca, and bc? Indicate whether the heat has gone into or out of the gas.
(d) Find the change in the internal energy of this hydrogen during segments ab, bc, and ca. Indicate whether the internal energy increased or decreased during each of these segments.
Answer:
Segment ab is isochoric, bc is isothermal, and ca is isobaric.
For bc, ΔT=0, ΔU=0 and Q=W=nRTln(VcVb)
For ideal H2 (diatomic), Cv=52R and Cp=72R
ΔU=nCvΔT for any process of an ideal gas.
(a) Tb=Tc. For states b and c, pV = nRT = constant so
pbVb=pcVc
Vc=Vb(pbpc)
Vc=(0.20 L)(2.0 atm0.50 atm)=0.80 L
(b) Ta=paVanR
Ta=(0.50 atm)(1.013×105 Pa/atm)(0.20×10−3 m3)(0.0040 mol)(8.3145 J/mol.K)
Ta=305 K
For states a and b, Va=Vb, so
Tp=nRV = constant, so we write
Tapa=Tbpb
Tb=Tc=Ta(pbpa)
Tb=(305 K)(2.0 atm0.50 atm)=1220 K=Tc
(c) For states a and b, Q = nCvΔT=n(52R)ΔT, which gives
Q=(0.0040 mol)(52(8.3145 J/mol.K))(1220 K−305 K)=+76 J
Q is positive and heat goes into the gas.
For states c and a, Q = nCpΔT=n(72R)ΔT, which gives
Q=(0.0040 mol)(72(8.3145 J/mol.K))(305 K−1220 K)=−107 J
Q is negative and heat comes out of the gas.
For states b and c:
Q=W=nRTln(VcVb), which gives
Q=(0.0040 mol)(8.3145 J/mol.K)(1200 K)ln(0.80 L0.20 L)=+56 J
Q is positive and heat goes into the gas.
(d) For states a and b:
ΔU=nCvΔT
ΔU=n(52R)ΔT, which gives
ΔU=(0.0040 mol)(52(8.3145 J/mol.K))(1220 K−305 K)=+76 J
The internal energy increased.
For states b and c:
ΔT = 0 so ΔU = 0. The internal energy does not change.
For states c and a:
ΔU=nCvΔT
ΔU=n(52R)ΔT, which gives
ΔU=(0.0040 mol)(52(8.3145 J/mol.K))(305 K−1220 K)=−76 J
The internal energy decreased.
The net internal energy change for the complete cycle a → b → c → d is ΔUtot=+76 J+0+(−76 J)=0.
For any complete cycle the final state is the same as the initial state and the net internal energy change is zero. For the cycle the net heat flow is
ΔQtot=+76 J+(−107 J)+56 J=+25 J.
The net work done in the cycle is positive and this agrees with our result that the net heat flow is positive.
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